Positive Integers - Difference of two squares

[Nazariy]

As you pointed out, \(\displaystyle a^2-b^2=(a+b)(a-b)\). If \(\displaystyle a,b\) are both odd or both even, then \(\displaystyle a+b\) and \(\displaystyle a-b\) are both even and so their product is divisible by 4. If one of \(\displaystyle a,b\) is odd and the other even, then \(\displaystyle a+b\) and \(\displaystyle a-b\) are both odd and so their product is odd. Hence \(\displaystyle a^2-b^2\) is always either odd or divisible by 4; it cannot be even and not divisible by 4.

This is perfect. Thank you.

 

[Ishuda]

\(\displaystyle 2=2*1
\\2=(a+b)(a-b)


\\ \begin{cases}
(a+b)=2,
\\(a-b)=1;
\end{cases}

\\ a=\frac{3}{2}
\\ b=\frac{1}{2}

\\ 2=\displaystyle{(\frac{3}{2})^2-(\frac{1}{2})^2}

\\ n=\frac{1}{2} \ and \ x=1
\)

n is a fraction and x=1, none of which is even

or

\(\displaystyle 2=2xn+x^2
\\(x+n)^2-n^2=2
\\x+n=\sqrt{n^2+2}
\\x=-n \pm \sqrt{n^2+2}
\)

Come now, don't change the rules. The question is meant to be talking about integer solutions otherwise chose a natural number n, any one of the uncountable numbers b<\(\displaystyle \sqrt{n}\) and let a=\(\displaystyle \sqrt{n-b^2}\) for a solution to the problem.

Nazariy, You apparently haven't really read the answers provided or, possibly, I haven't been able to explain myself well enough. In either case, I think I'm finished with this problem.

 

[Nazariy]

Come now, don't change the rules. The question is meant to be talking about integer solutions otherwise chose a natural number n, any one of the uncountable numbers b<\(\displaystyle \sqrt{n}\) and let a=\(\displaystyle \sqrt{n-b^2}\) for a solution to the problem.

Nazariy, You apparently haven't really read the answers provided or, possibly, I haven't been able to explain myself well enough. In either case, I think I'm finished with this problem.

But there are no integer number solutions. I have read the answers.

thank you for your help, you have answered my question anyway